The usual suspects in this subject (Borwein, Bailey, etc.) have written good expositions of this topic. What follows is shamelessly stolen from them. An easier example to explain, that illustrates the main idea, is log 2. We have the formula
$$\log 2 = \sum_{n=1}^\infty {1 \over n 2^n}.$$
You can, if you like, call this a "BBP formula for log 2" since it lets you calculate binary digits of log 2 beginning after position $d$ as follows:
$$\eqalign{2^d \log 2 \bmod 1 &= \sum_{n=1}^d {2^{d-n}\over n} \bmod 1 + \sum_{n=d+1}^\infty {2^{d-n}\over n}\cr &= \biggl(\sum_{n=1}^d {2^{d-n}\bmod n\over n}
\biggr) \bmod 1 + \sum_{n=d+1}^\infty {2^{d-n}\over n}}.$$
The main point is that the second term here is small while the first term—call it $x_d$—is readily computable, and in fact satisfies the recurrence
$$x_d = \biggl(2x_{d-1} + {1\over d}\biggr)\bmod 1.$$
So if one can demonstrate that the sequence $(x_d)$ defined by the above recurrence
is equidistributed in the unit interval, then one can conclude (after fussing a little with the error term) that log 2 is normal in base 2.

For $\pi$, the same argument goes through, except that using the (original) BBP formula for $\pi$, one is led to consider a slightly different recurrence, with $2x_{d-1}$ replaced by $16x_{d-1}$ and $1/d$ replaced by a more complicated rational function of $d$.

The upshot is that BBP-type formulas allow us to rephrase the normality of certain constants in terms of the equidistribution of certain simple-looking recurrences. But currently, nobody has any idea how to prove equidistribution. Thus I would not say that there are "difficulties in using the BBP formula for the purpose of proving normality"; rather, the BBP formula can be applied straightforwardly, but the basic difficulty remains intact.